Vertex stabilizers of locally s-arc transitive graphs of pushing up type

Abstract

Suppose that $\Delta$ is a thick, locally finite and locally s-arc transitive G-graph with $s \ge 4$. For a vertex z in $\Delta$, let $G_z$ be the stabilizer of z and $G_z^{[1]}$ the kernel of the action of $G_z$ on the neighbours of z. We say $\Delta$ is of pushing up type provided there exist a prime p and a 1-arc (x, y) such that $C_{G_z}(O_p(G_z^{[1]})) \le O_p(G_z^{[1]})$ for $z \in \{x,y\}$ and $O_p(G_x^{[1]}) \le O_p(G_y^{[1]})$. We show that if $\Delta$ is of pushing up type, then $O_p(G_x^{[1]})$ is elementary abelian and $G_x/G_x^{[1]}\cong X$ with $\textrm{PSL}_2(p^a)\le X \le \mathrm{P\Gamma L}_2(p^a)$.

1 Introduction

In this article, we consider graphs $\Delta$ that are connected, undirected and without loops or multiple edges. The vertex set of $\Delta$ is denoted by $V\Delta$, and the edge set is $E\Delta$. A G-graph is a graph $\Delta$ together with a subgroup $G \le \textrm{Aut}(\Delta )$. An s-arc emanating from $x_0\in V\Delta$ is a path $(x_0, x_1, \dots ,x_s)$ with $x_{i-1}\ne x_{i+1}$ for $1\le i \le s-1$. Denote by $G_z$ the stabilizer of a vertex $z\in V\Delta$.

A G-graph $\Delta$ is

• Thick if the valency at each vertex is at least 3;

• Locally finite if for each $z \in V\Delta$, $G_z$ is a finite group;

• Locally s-arc transitive if for every vertex $z \in V\Delta$, $G_z$ is transitive on the set of s-arcs emanating from z.

This paper is part of ongoing research aimed at determining all vertex stabilizer amalgams for thick, locally finite and locally s-arc transitive G-graphs for $s\ge 4$. Throughout this introduction, $\Delta$ represents a thick, locally finite G-graph.

It is easy to see that if $\Delta$ is a locally s-arc transitive G-graph with $s\ge 1$, then G is transitive on $E \Delta$ and thus G has at most two orbits on $V \Delta$. If $s\ge 2$, then, for $x \in V \Delta$, we also know that $G_x$ acts 2-transitively on $\Delta (x)=\{v\mid \{x,v\}\in E\Delta \}$. For $(x_1,x_2)$ a 1-arc in $\Delta$, the triple $(G_{x_1},G_{x_2};G_{x_1}\cap G_{x_2})$ is called the vertex stabilizer amalgam of $\Delta$ with respect to the 1-arc $(x_1,x_2)$. When we study s-arc transitive G-graphs with $s\ge 1$, it is impossible to determine $\Delta$. The best we can hope for is a description of the vertex stabilizer amalgam, and the best of all this will be described up to isomorphism of the amalgam.

For a vertex $z \in V\Delta$, $G_z$ acts on $\Delta (z)$. The kernel of this action is denoted by $G_z^{[1]}$, and $G_z^{\Delta (z)}$ is the permutation group $G_z/G_z^{[1]}$. A locally finite and locally G-graph $\Delta$ is of

• Local characteristic p, if there exists a prime p such that

  $$\begin{aligned} C_{G_z} (O_p(G_z ^{[1]})) \le O_p(G_z ^{[1]}) \hbox {, for all } z \in V\Delta ; \end{aligned}$$

• Pushing up type with respect to the 1-arc (x, y) and the prime p, if $\Delta$ is of local characteristic p and

  $$\begin{aligned} O_p(G_{x} ^{[1]}) \le O_p(G_{y} ^{[1]}). \end{aligned}$$

Assume that (x, y) is a 1-arc and set $G_{x,y}= G_x \cap G_y$. One consequence of the local characteristic p property is that $O_p(G_x^{[1]})$ and $O_p(G_y^{[1]})$ are non-trivial. In particular, this means that $G_x$ and $G_y$ are rather large and have potentially complicated structure. Notice that if K is a subgroup of $G_{x,y}$, and K is normalized by both $G_{x}$ and $G_y$, then K fixes every vertex of $\Delta$ and is consequently trivial. Hence, if $\Delta$ is of pushing up type with respect to (x, y), then, as $O_p(G_x^{[1]})$ is non-trivial, we learn that $\Delta$ is not of pushing up type with respect to (y, x). Hence, in these circumstances, G has two orbits on $V\Delta$.

The generic examples of thick, locally finite and locally s-arc transitive G-graphs $\Delta$ with $s \ge 4$ have vertex stabilizer amalgams which are weak BN-pairs [6]. One of the achievements in [1, Theorem 1] is the proof that, for $s \ge 6$, the generic examples are the only examples. In particular, [1, Corollary 1] remarks that $s \le 9$ for any such G-graph $\Delta$. In [5], examples of G-graphs which are of pushing up type with $s = 5$ have been constructed via amalgams in $\textrm{Sym}(p^{2a})$. Thus, the vertex stabilizer amalgam of $\Delta$ may not to be a weak BN-pair when $s \le 5$. To determine the G-graphs $\Delta$ which are thick, locally finite and locally s-arc transitive with $4\le s \le 5$, as in [1], we consider three distinct cases:

• $\Delta$ is not of local characteristic p.

• $\Delta$ is of local characteristic p but not of pushing up type.

• $\Delta$ is of pushing up type.

In the first case, [2, Theorem 1] shows that $s=5$ and the vertex stabilizer amalgams are either isomorphic to certain subamalgams of the vertex stabilizer amalgam of the G-graph for $p=2$ constructed in [7], or isomorphic to the amalgam of two maximal p-local subgroups of $\textrm{Aut}({}^3\mathrm D_4(2))$ (for $p = 7$) or of $\textrm{Aut}(\mathrm J_2)$ (for $p = 5$).

We expect that the second possibility yields weak BN-pairs. The configurations appearing in option three are the subject of this article together with its companions [3, 4] which are in preparation.

Assume from now on that $\Delta$ is of pushing up type with respect to (x, y) and the prime p. For $z \in V\Delta$, set $Q_z=O_p(G_z^{[1]})$ and

$$\begin{aligned} L_z = \langle Q_u \mid u \in \Delta (z) \rangle Q_z. \end{aligned}$$

The main result of this paper is as follows.

Theorem 1.1

Suppose that $s \ge 4$ and $\Delta$ is a thick, locally finite, locally s-arc transitive G-graph of pushing up type with respect to the 1-arc (x, y) and the prime p. Then p is odd and the following hold

• (a) $G_x^{\Delta (x)} \cong X$ where $\textrm{PSL}_2(p^a)\le X \le \mathrm{P\Gamma L}_2(p^a)$ and $\Delta (x)$ has size $p^a+1$ and can be identified with the projective line for X;

• (b) $L_x/Q_x \cong \textrm{SL}_2(p^a)$, $O^p(L_x) \cong \textrm{ASL}_2(p^a)^\prime$ and $Q_x$ is an elementary abelian p-group.

In [3, 4], van Bon establishes the isomorphism types of the vertex stabilizer amalgams appearing in the conclusion of Theorem 1.1 and so it completes the determination of G-graphs of pushing up type with $s \ge 4$. This then extends [1, Lemma 7.7] which can be interpreted to say that if $\Delta$ is of pushing up type, then $s\le 5$.

The organization of this paper is as follows. In Sect. 2, we derive general properties of vertex stabilizer amalgams for $\Delta$ of pushing up type. In Sect. 3, we consider the possibility that $G_x^{\Delta (x)}$ is a projective linear group of degree at least 3; the main result of the section is Proposition 3.1 which asserts that $F^*(G_x/G_x^{[1]}) \not \cong \textrm{PSL}_n(p^a)$ with $n \ge 3$. The strategy followed to obtain the conclusion of Proposition 3.1 uses the results of Sect. 2 and is similar in flavour to [2] which exploits Zsigmondy primes. This fails to eliminate the possibility that $F^*(G_x/G_x^{[1]}) \cong \textrm{PSL}_3(2)$, however, and so here we call upon a pushing up result [10] which allows us to compare non-central chief factors of the vertex stabilizers. This eventually leads to the elimination of this last case as well. Finally, in Sect. 4, we recall that the action of $G_x^{\Delta (x)}$ on $\Delta (x)$ is 2-transitive, and so with the help of the classification of finite 2-transitive groups and Proposition 3.1, we see that $L_x$ is a rank 1 Lie type group or is of regular type. This is precisely the situation handled in Sect. 3 of [1]. After application of these results to our case, we are left in a situation where we can follow steps 1–9 of the proof of [1, Lemma 7.7] word for word to obtain the theorem.

Throughout this paper, we assume the following hypothesis:

MainHypothesis

The G-graph $\Delta$ is thick, locally finite, locally s-arc transitive with $s \ge 4$ and, in addition, is of pushing up type with respect to the 1-arc (x, y) and prime p.

The notation used in the paper is standard in the theory of (locally) s-arc transitive G-graphs and given in Sect. 2. Our group theoretic notation follows [9].

2 Preliminaries

In this section, we prove some properties of thick, locally finite and locally s-arc transitive G-graphs. We assume the Main Hypothesis, though some results also hold under weaker assumptions. First we fix the notation used throughout the article.

Notation 2.1

Let $d(\cdot , \cdot )$ represent the standard distance function on $\Delta$. For $u \in V\Delta$, (u, v) a 1-arc in $\Delta$, $\Theta \subseteq V \Delta$ and $i \ge 1$,

$$\begin{aligned} \Delta ^{i} (u)= & {} \{v \in V\Delta \mid d(u,v)\le i\};\\ \Delta (u)= & {} \Delta ^1(u) \backslash \{u\};\\ q_u= & {} |\Delta (u)|-1;\\ G_u= & {} \{ g\in G \mid u^g=u \}; \\ G_\Theta= & {} \bigcap _{\theta \in \Theta } G_\theta ;\\ G_u^{[i]}= & {} G_{\Delta ^i(u)};\\ G_{\Theta }^{[1]}= & {} \bigcap _{\theta \in \Theta } G_\theta ^{[1]};\\ Q_u= & {} O_p(G_u^{[1]});\\ Z_u= & {} \Omega _1(Z(Q_u));\\ C_u= & {} \langle G_v^{[2]}\mid v \in \Delta (u)\rangle Q_u;\\ L_u= & {} \langle Q_v \mid v\in \Delta (u) \rangle Q_u;\\ L_{u,v}= & {} G_{u,v} \cap L_u; \end{aligned}$$

and (w, x, y, z) is a fixed 3-arc where (x, y) is an arc for which $\Delta$ is of pushing up type.

Notice that we do not know that $L_{u,v} = L_{v,u}$, and so the order of the vertices on the arc is important for the definition of $L_{u,v}$. We recall the fundamental properties of the G-graph $\Delta$:

$$\begin{aligned}C_{G_u}(Q_u)\le & {} Q_u\text { for } u \in \{x,y\}; \text { and }\\ Q_x\le & {} Q_y. \end{aligned}$$

An essential tool when studying such G-graphs is given by the following lemma:

Lemma 2.2

Assume that $R\le G_{x,y}$, $N_{G_x}(R)$ is transitive on $\Delta (x)$ and $N_{G_y}(R)$ is transitive on $\Delta (y)$. Then $R=1$.

Proof

See [9, 10.3.3]. $\square$

Because $\Delta$ has pushing up type, $Q_x \le Q_y$. In fact, $Q_x< Q_y$ as otherwise $Q_x=Q_y=1$ by Lemma 2.2, and then, $\Delta$ is not of local characteristic p. One consequence of this observation, as mentioned in introduction, is that G has two orbits on $\Delta$.

We present some elementary properties of the subgroups defined in Notation 2.1.

Lemma 2.3

Let $N_x=O^p(L_{x})$, and $\widetilde{G}_{x}=G_{x}/Q_{x}$. Then

• (i) $[ G_{x}^{[1]},L_{x}]\le Q_{x}$;

• (ii) $N_x \not \le G_{x}^{[1]}$;

• (iii) $L_{x}= N_xQ_{y}$ and $G_x=N_xG_{x,y}$;

• (iv) either $\widetilde{N_x}$ is quasisimple or an r-group, r a prime with $r\ne p$;

• (v) $\widetilde{G_x^{[1]}\cap L_x}=\Phi (\widetilde{L}_{x})=Z(\widetilde{L}_{x})$;

• (vi) if $\widetilde{N_x}$ is an r-group, then $\widetilde{G_x^{[1]}\cap L_x}=\widetilde{N_x}^\prime = \Phi (\widetilde{N_x})$, and $G_{x}$ acts transitively on the non-trivial elements of $N_x^{\Delta (x)}$.

Proof

The first statement follows from [1, Lemma 5.1 (a)] while the remainder of the statements can be found in [1, Lemma 5.2]. $\square$

One important and frequently used consequence of Lemma 2.3 (iii) is that $L_x$ operates transitively on $\Delta (x)$.

Lemma 2.4

The following hold:

• (i) $G_{x,y}^{[1]}=G_x^{[2]}$, $G_{y}^{[2]}=G_y^{[3]}$ and $Q_x=O_p(G_x^{[2]})$;

• (ii) $G_{x} ^{[1]} \cap G_{z} = G_{x}^{[2]}$.

In particular, $G_{x} ^{[1]}$ induces a semi-regular group on $\Delta (y) \backslash \{{x}\}$, $|G_{x}^{[1]}G_y^{[1]}/G_{y}^{[1]}|$ divides $q_y$ and $|G_{x,y,z}| = |G_{x,y,z} ^{\Delta (x)}||G_{x} ^{[2]}|.$

Proof

(i). By Lemma 2.3 (i) and (iii), $[L_x,G_x^{[1]}] \le Q_x$ and $L_x$ acts transitively on $\Delta (x)$. Hence,

$$\begin{aligned}{}[L_x,G_{x,y}^{[1]}]\le [L_x,G_x^{[1]}] \le Q_x \le G_{x,y}^{[1]}. \end{aligned}$$

Thus, $L_x$ normalizes $G_{x,y}^{[1]}$. The transitivity of $L_x$ on $\Delta (x)$ now yields $G_{x,y}^{[1]}=G_x^{[2]}$. In particular, $G_y^{[2]} \le G_{u,y}^{[1]}=G_u^{[2]}$ for any $u\in \Delta (y)$. It follows that $G_{y}^{[2]}=G_y^{[3]}$. Since $Q_x \le Q_v \le G_v^{[1]}$ for all $v \in \Delta (x)$, we also have $Q_x \trianglelefteq G_x^{[2]}$. Hence, $Q_x=O_p(G_x^{[2]})$.

(ii). By Lemma 2.3 (i) and, as the G-graph is pushing up type,

$$\begin{aligned}{}[L_x, G_x^{[1]}\cap G_z]\le [L_x, G_x^{[1]}] \le Q_x \le G_x^{[1]}\cap Q_y\le G_x^{[1]}\cap G_z. \end{aligned}$$

Thus, $L_x$ normalizes $G_x^{[1]}\cap G_z$. Pick $g \in L_x$ with $y^g \not = y$, and put $y^\prime = y^g$ and $z^\prime = z^g$. Then $G_x^{[1]}\cap G_z= G_x^{[1]}\cap G_{z^\prime }$. Since $s\ge 4$, $G_{{z^\prime },{y^\prime },x,y}$ acts transitively on $\Delta (y) \backslash \{x\}$ and it also normalizes $G_x^{[1]}\cap G_{z^\prime }= G_x^{[1]}\cap G_z$. Therefore, $G_x^{[1]} \cap G_{z} \le G_{y} ^{[1]}$. Hence, from (i),

$$\begin{aligned} G_x^{[2]} \le G_x^{[1]}\cap G_z \le G_{x,y} ^{[1]}=G_x ^{[2]}. \end{aligned}$$

Now (ii) follows.

Finally, as $(G_{x} ^{[1]} \cap G_{z})^{\Delta (y)} = 1$ by (ii), $G_{x} ^{[1]}$ induces a group which acts semi-regularly on $\Delta ({y}) \backslash \{x\}$. Therefore, we also have $|G_{x} ^{[1]}G_y^{[1]}/G_y^{[1]}|$ divides $\Delta (y)-1=q_y$. To see that $|G_{x,y,z}| = |G_{x,y,z} ^{\Delta (x)}||G_{x} ^{[2]}|,$ just observe that (ii) gives

$$\begin{aligned} G_x^{[2]} \le G_{x}^{[1]} \cap G_{x,y,z} \le G_{x}^{[1]}\cap G_z = G_{x}^{[2]}. \end{aligned}$$

$\square$

Lemma 2.5

The following hold:

• (i) $G_y^{[2]}$ and $G_x^{[1]} \cap G_z ^{[1]}$ are p-groups;

• (ii) $G_x^{[1]} \cap G_z ^{[1]} =G_x^{[2]} \cap G_z ^{[2]}= G_{x,y,z}^{[1]}=Q_x \cap Q_z$;

• (iii) $Q_yG_x^{[1]} \cap G_z^{[2]}=Q_z$;

• (iv) $Q_yG_{x}^{[2]} \cap Q_yG_{z}^{[2]} = Q_y$;

• (v) either $G_x^{[2]}/Q_x$ is abelian or $G_x^{[1]}=G_x^{[2]}$;

• (vi) ${\textrm{lcm}}(q_x,q_y)$ divides $|G_{w,x,y,z}|$ and $\pi (G_{x,y})=\pi (G_{w,x,y,z})$.

Proof

(i). Let $R \in {\textrm{Syl}}_r (G_x^{[1]})$, with $r \ne p$. Then, Lemma 2.3 (i) yields $[R,L_x] \le [G_x^{[1]},L_x] \le Q_x$. Hence, $L_x$ normalizes $Q_xR$ and acts on $Q_xR$ by conjugation. Therefore, the Frattini argument gives $L_x= Q_xN_{L_x}(R)$. In particular, $N_{L_x}(R)$ is transitive on $\Delta (x)$. Now using Lemma 2.3 (i) we get $[R,N_{L_x}(R)] \le [G_x^{[1]},L_x]\cap R\le Q_x \cap R=1$. So for all $R_0 \le R$, $C_{L_x}(R_0)\ge C_{L_x}(R)=N_{L_x}(R)$ is transitive on $\Delta (x)$.

Let $T \in {\textrm{Syl}}_t (G_y^{[2]})$, with $t \ne p$. By the Frattini argument $G_y= N_{G_y}(T)G_y^{[2]}$. Hence, $N_{G_y}(T)$ is transitive on $\Delta (y)$. Since $T \le G_x^{[1]}$, we have $C_{L_x}(T)$ is transitive on $\Delta (x)$. Thus, T is normalized by both $N_{G_y}(T)$ and $C_{L_x}(T)$ and so $T=1$ by Lemma 2.2. We conclude that $G_y^{[2]}$ is a p-group as claimed.

Suppose that $T\in {\textrm{Syl}}_t(G_x^{[1]} \cap G_z^{[1]})$, with $t \ne p$. Since $s \ge 4$, there exists a $c \in C_{L_x}(T)$ with $y^c \ne y$ and $z^c \ne z$. Indeed, that we can choose c with $y^c\ne y$ follows from the transitivity of $G_x$ on $\Delta (x)$. We claim for such a $c\in C_{L_x}(T)$, $z^c \ne z$. In the counter case, $(x,y,z,y^c,x)$ is a circuit of length 4 and $G_{x,y,z,y^c}$ is transitive on $\Delta (y^c) {\setminus } \{z\}$ as $s \ge 4$. As $x\in \Delta (y^c)$, we infer that $|\Delta (y^c)|=2$, contrary to $\Delta$ being thick.

Let $\gamma = (z^c,y^c,x, y)$ and $N=G_x^{[1]} \cap G_{z^c}^{[1]}$. Now $T=T^c \in {\textrm{Syl}}_t(N)$ and $N\trianglelefteq G_{\gamma }$. Then, by Lemma 2.4 (ii), $N \le G_x^{[2]}\le G_y^{[1]}$. Since $s \ge 4$, $G_{\gamma }$ is transitive on $\Delta (y) \backslash \{x\}$. By the Frattini argument $G_{\gamma }= N_{G_{\gamma }}(T)N$. Thus, $N_{G_{\gamma }}(T)$ is transitive on $\Delta (y) \backslash \{x\}$ since $N \le G_y^{[1]}$. As $T\le G_{z}^{[1]}$, $T \le G_u^{[1]}$, for all $u \in \Delta (y)$. But then, $T \le G_y^{[2]}$ which is a p-group and thus $T=1$. This completes the proof of (i).

(ii). From Lemma 2.4 (ii), we have $G_x^{[1]} \cap G_z =G_x^{[2]}$ and $G_x \cap G_z^{[1]}=G_z^{[2]}$, and thus,

$$\begin{aligned} G_x^{[2]} \cap G_z^{[2]} \le G_{x,y,z}^{[1]} \le G_x^{[1]}\cap G_z^{[1]} \le G_x^{[2]} \cap G_z^{[2]} \end{aligned}$$

which gives $G_x^{[1]}\cap G_z^{[1]}= G_{x,y,z}^{[1]}=G_x^{[2]} \cap G_z^{[2]}$.

Since, by (i), $G_x^{[1]} \cap G_z^{[1]}=G_x^{[2]} \cap G_z^{[2]}$ is a p-group which is normalized by $G_x^{[2]}$ and $G_z^{[2]}$, Lemma 2.4 (i) implies $G_x^{[1]} \cap G_z^{[1]} \le O_p (G_x^{[2]})=Q_x$ and $G_x^{[1]} \cap G_z^{[1]} \le O_p (G_z^{[2]})= Q_z$. Hence

$$\begin{aligned} Q_x \cap Q_z \le G_x^{[2]} \cap G_z^{[2]} \le Q_x \cap Q_z \end{aligned}$$

and this completes the proof of (ii).

(iii). Suppose that $R\le Q_yG_x^{[1]} \cap G_z^{[2]}$ has $p'$-order. Since $G_x^{[1]}$ is normalized by $Q_y$, we have $R \le G_x^{[1]}$. Hence, $R \le G_x^{[1]} \cap G_z^{[2]}$ which by (i) is a p-group. Thus, $R=1$ and $Q_yG_x^{[1]} \cap G_z^{[2]}$ is a p-group. Since $Q_z \le Q_yG_x^{[1]} \cap G_z^{[2]} \trianglelefteq G_z^{[2]}$ and $O_p(G_z^{[2]})=Q_z$, we have $Q_yG_x^{[1]} \cap G_z^{[2]}=Q_z$.

(iv). Using (iii) and the modular law, we have

$$\begin{aligned} Q_yG_x^{[2]} \cap Q_yG_z^{[2]}=Q_y(Q_yG_x^{[2]}\cap G_{z}^{[2]}) \le Q_y(Q_yG_x^{[1]} \cap G_z^{[2]})=Q_yQ_z=Q_y. \end{aligned}$$

(v). Suppose $G_x^{[1]} \ne G_x^{[2]}$. Let $g \in G_x^{[1]} \setminus G_x^{[2]}$. Then, by Lemma 2.4 (ii), $z^g \ne z$. Then, $G_x^{[1]}G_z^{[2]}=(G_x^{[1]}G_z^{[2]})^g=G_x^{[1]}G_{z^g}^{[2]}$. Hence, using (ii) and the modular law, we have

$$\begin{aligned}{}[G_z^{[2]},G_z^{[2]}]\le & {} [G_x^{[1]}G_z^{[2]},G_x^{[1]}G_z^{[2]}] \cap G_z^{[2]}\le [G_x^{[1]}G_z^{[2]},G_x^{[1]}G_{z^g}^{[2]}]\cap G_z^{[2]}\\\le & {} G_{x}^{[1]}[G_z^{[2]},G_{z^g}^{[2]}]\cap G_z^{[2]}\le G_{x}^{[1]}(G_z^{[2]} \cap G_{z^g}^{[2]})\cap G_z^{[2]} \\\le & {} G_{x}^{[1]} Q_z\cap G_z^{[2]}= Q_z. \end{aligned}$$

It follows that $G_z^{[2]}/Q_z$ is abelian.

(vi). Since $s \ge 4$, $|G_{w,x,y,z}|$ is divisible by ${{\,\textrm{lcm}\,}}(q_x,q_y)$. Let $t \in \pi (G_{x,y})$ and $h \in G_{x,y}$ have order t. If t divides $q_xq_y$, then t divides $|G_{w,x,y,z}|$. Suppose t does not divide $q_xq_y$. Then, $\langle h\rangle$ fixes a vertex in both $u^\prime \in \Delta (x) \backslash \{y\}$ and $z^\prime \in \Delta (y) \backslash \{x\}$. Thus, t divides $|G_{u^\prime ,x,y,z^\prime }|=|G_{w,x,y,z}|$ since $s \ge 4$. It follows that $\pi (G_{x,y}) \subseteq \pi (G_{w,x,y,z})$. The reverse inclusion is immediate since $G_{w,x,y,z}$ is a subgroup of $G_{x,y}$. $\square$

Lemma 2.6

If $G_x^{[2]}/Q_x$ is abelian of exponent t, then $C_y/Q_y$ is abelian of exponent t and has order at least $t^2$.

Proof

For $u,v \in \Delta (y)$, $[G_u^{[2]},G_v^{[2]}] \le Q_u\cap Q_v\le Q_y$ by Lemma 2.5 (ii) and the fact that $G_x^{[2]}/Q_x$ is abelian. Hence, $C_y/Q_y$ is abelian and, as $G_x^{[2]}Q_y\cap G_u^{[2]}Q_y= Q_y$ by Lemma 2.5 (iv), the claim follows. $\square$

The proof of the following lemma is based on an argument that can be found in [1, Lemma 4.8].

Lemma 2.7

Let t be a prime dividing $q_x$. Then $G_x^{[2]}$ is a $t^\prime$-group if and only if $G_x^{[1]}$ is a $t^\prime$-group.

Proof

It suffices to prove that $G_{x}^{[1]}$ is a $t'$-group whenever $G_{x}^{[2]}$ is a $t'$-group. Assume that $G_{x}^{[2]}$ is a $t'$-group. Let (x, y, z, a) be a 3-arc and let $T \in {\textrm{Syl}}_t(G_{x,y,z,a})$. Then, as $s\ge 4$ and t divides $q_x$, T acts non-trivially on $\Delta (x)$. Hence, again as $s \ge 4$, T fixes no 4-arcs starting with a vertex in $a^G$. If $N_{G_y}(T) \not \le G_z$, then there exists $g\in N_{G_y}(T){\setminus }{G_z}$ such that $T=T^g \le G_{x,y,z,a}^g = G_{x^g,y,z^g,a^g}$. Hence, T fixes the 4-arc $(a,z,y,z^g,a^g)$, which is a contradiction. Therefore, $N_{G_y}(T)\le G_z$. Let $X \in {\textrm{Syl}}_t(G_{x}^{[1]})$ be normalized by T. Then, $N_{X}(T)\le G_z$. Hence,

$$\begin{aligned} N_{X}(T) \le X \cap G_z\le G_{x}^{[1]} \cap G_z = G_x^{[2]} \end{aligned}$$

by Lemma 2.4 (ii) and so $N_X(T)=1$ as $G_x^{[2]}$ is a $t'$-group. It follows that $C_{X}(T) \le N_X(T)=1$, and so we conclude that $X=1$. Hence, $G_x^{[1]}$ is a $t'$-group and this establishes the claim. $\square$

The next lemma, which is fundamental for the approach to our proof of Theorem 1.1, is the origin of the name pushing up type amalgam.

Lemma 2.8

Suppose that K is a non-trivial characteristic subgroup of $Q_y$. Then, $N_{G_{x}}(K)= G_{x,y}$. In particular, no non-trivial characteristic subgroup of $Q_y$ is normalized by $L_x$.

Proof

Suppose K is a non-trivial characteristic subgroup of $Q_y$. Then, K is normalized by $G_y$ and so also by $G_{x,y}$. Since $G_x$ acts 2-transitively on $\Delta (x)$, $G_{x,y}$ is a maximal subgroup of $G_x$. Hence, if $N_{G_x}(K) > G_{x,y}$, then K is normalized by $G_x$. But then K is normalized by $G_x$ and $G_y$ and so $K=1$ by Lemma 2.2. $\square$

3 $L_x^{\Delta (x)}$ is not a projective linear group of degree at least 3

In this section, we intend to demonstrate

Proposition 3.1

Suppose that $s\ge 4$ and $\Delta$ is a thick, locally finite, locally s-arc transitive G-graph of pushing up type with respect to (x, y) and the prime p. Then, for $n \ge 3$ and a a natural number, $F^*(G_x^{\Delta (x)}) \not \cong \textrm{PSL}_n(p^a)$ acting on projective points.

Throughout this section, we assume that the Main Hypothesis holds and

$$\begin{aligned} L_x ^{\Delta (x)} \cong \textrm{PSL}_n(q) \end{aligned}$$

with $n \ge 3$, $q=p^a$ and $\Delta (x)$ corresponding to the points in projective $(n-1)$-space. We continue with the notation established in Notation 2.1.

Before we start on the proof, we record the following facts about projective linear groups.

Lemma 3.2

Assume that $n \ge 3$, p is a prime, $q=p^k$ and $\textrm{PSL}_n(q) \trianglelefteq H \le \mathrm{P\Gamma L}_n(q)$ acting on the projective space PV. Let $u,v \in PV$ be distinct points. The following statements hold:

• (i) $|H/H_v|-1= q\left( \frac{q^{n-1}-1}{q-1}\right)$.

• (ii) There exists a unique $E\trianglelefteq H_{v}$ such that $O_p(E)=O_p(H_v)$, $E/O_p(E) \cong \textrm{SL}_{n-1}(q)$ and $|H_v/ E|$ divides $(q-1)k$. Moreover, $O_p(E)$ is a natural module for $E/O_p(E)$.

• (iii) Either $n=3$ and $|H_{u,v}/O_p( H_{u,v})|$ divides $(q-1)^2k$, or there exists $O_p(H_{u,v}) \trianglelefteq F\trianglelefteq H_{u,v}$ such that $F/O_p(F) \cong \textrm{SL}_{n-2}(q)$ and $|H_{u,v}/F|$ divides $(q-1)^2k$.

• (iv) Let N and E be subgroups of $H_v$ with $O_p(H_v)=O_p(N)=O_p(E)$ and $E/O_p(E) \cong \textrm{SL}_{n-1}(q)$. Suppose E normalizes N. Then, one of the following holds:

  • (a) $E \le N$;

  • (b) $N/O_p(H_v)$ cyclic, $[N,E] \le O_p(H_v)$ and $|N/O_p(H_v)|$ divides $(q-1)$;

  • (c) $E^\prime \le N < E$, $n=3$ and either

    • (ci) $q=p=2$ and $N/O_2(H_v) \cong {\mathrm C}_3$; or

    • (cii) $q=p=3$ and $N/O_3(H_v) \cong {\mathrm Q}_8$.

Proof

$(i)-(iii)$. Straight forward.

(iv). For a subgroup $X \le H_v$, we will denote with $\overline{X}$ its image in $H_v/O_p(H_v)$. Since E normalizes N, we have $[\overline{N}, \overline{E}] \trianglelefteq \overline{E}$. Hence either $\overline{E} \le [\overline{N},\overline{E}]$, $[\overline{N},\overline{E}] \le Z(\overline{E})$, or $n=3=q$ and $[\overline{N}, \overline{E}] = [\overline{E}, \overline{E}] \cong Q_8$ or $n=3$, $q=2$ and $[\overline{N}, \overline{E}]= [\overline{E}, \overline{E}] \cong C_3$. In the second case, the Three Subgroup Lemma gives $[\overline{N},[\overline{E}, \overline{E}]]=1$. It follows that $\overline{N} \le C_{\overline{H}_v}( [\overline{E}, \overline{E}])$. Thus $[\overline{N}, \overline{E}]=1$ too and $\overline{N}$ is a cyclic group whose order divides $q-1$. In the third case, $\overline{E} \le \overline{N}$ or $\overline{N} \cong {\mathrm Q}_8$. In the fourth case, $\overline{E} \le \overline{N}$ or $\overline{N} \cong {\mathrm C}_3$. $\square$

We begin the proof of Proposition 3.1 with a lemma which restricts the structure of $G_x^{[2]}/Q_x$.

Lemma 3.3

One of the following holds:

• (i) $G_x^{[2]}/Q_x \cong {\mathrm C_t}$ with t dividing $q-1$; or

• (ii) $G_x ^{\Delta (x)} \cong \textrm{PSL}_3(2)$ and $G_x^{[2]}/Q_x \cong {\mathrm C}_3$.

Proof

Let $P=G_{x,y}$ and put $\overline{P} = P/G_{x}^{[1]}Q_y$. Then, as $G_x^{[1]}Q_y= G_x^{[1]}O_p(P)$, Lemma 3.2 implies $\textrm{SL}_{n-1}(q) \trianglelefteq \overline{P} \le \Gamma {\mathrm L}_{n-1}(q)$. Let $\overline{E} \le \overline{P}$ with $\overline{E}\cong \textrm{SL}_{n-1}(q)$. Recall that $C_y \le G_y^{[1]}$ and $C_y$ is normal in $G_y$. Hence, $\overline{C}_y$ is a normal subgroup of $\overline{P}$.

Let $u \in \Delta (y){\setminus }\{x\}$ with $z\ne u$. Then, by Lemma 2.5 (iv), $[G_{z}^{[2]},G_{u}^{[2]}]\le Q_y$. It follows that $Q_yG_z^{[2]}$ is normal in $C_y$. Since $\overline{G_z^{[2]}}$ is normal in $\overline{C_y}$, $\overline{G_z^{[2]}}$ is subnormal in $\overline{P}$. If $\overline{G_z^{[2]}}$ does not centralize $\overline{E}$, then $\overline{E}'\le \overline{G_z^{[2]}}$ by Lemma 3.2 (iii) and Lemma 2.6. Since this is true for all $z \in \Delta (y){\setminus }\{x\}$ and $\overline{[G_{z}^{[2]},G_{u}^{[2]}]}=1$, this is impossible unless $(n,q)=(3,2)$ and $\overline{G_z^{[2]}}$ is cyclic of order 3. Hence, in the general case, $\overline{G_z^{[2]}}$ centralizes $\overline{E}$ and we conclude that $\overline{G_z^{[2]}}$ is cyclic of order t dividing $q-1$. We have demonstrated

$$\begin{aligned} \overline{G_z^{[2]}}\cong {\left\{ \begin{array}{ll}{\mathrm C}_t&{} t \text { divides } q-1\\ \textrm{C}_3&{}(n,q)= (3,2).\end{array}\right. } \end{aligned}$$

Finally, Lemma 2.5 (iii) yields

$$\begin{aligned} G_z^{[2]}/Q_z=G_z^{[2]} /(G_z^{[2]}\cap G_x^{[1]}Q_y)\cong G_z^{[2]}G_{x}^{[1]}Q_y/G_{x}^{[1]}Q_y=\overline{G_z^{[2]}} \end{aligned}$$

and this completes the proof.$\square$

Lemma 3.4

If $F^*(G_x ^{\Delta (x)}) \cong \textrm{PSL}_n(q)$ with $n \ge 3$, then $F^*(G_x ^{\Delta (x)}) \cong \textrm{PSL}_3(2)$ and $G_x^{[2]}/Q_x \cong {\mathrm C}_3$.

Proof

Suppose that $F^*(G_x ^{\Delta (x)}) \not \cong \textrm{PSL}_3(2)$ with $G_x^{[2]}/Q_x \cong {\mathrm C}_3$. We first prove that

${\textbf{1}}^\circ$. $q^{n-1} = 2^6$ or $q^{n-1}-1 = p^2-1$ and p is a Mersenne prime.

Suppose that t is a Zsigmondy prime for $((n-1)k,p)$. By [2, 3.8], t does not divide $(n-1)k$, and since $n-1 \ge 2$, t does not divide $q-1$. Therefore, as t divides $q_x=q(q^{n-1}-1)/(q-1)$, combining Lemmas 2.7 and 3.3 yields t does not divide $|G_x^{[1]}|$. Since t divides $|G_{x,y}|$, t divides $|G_{w,x,y}|$ by Lemma 2.5 (vi). Therefore, t divides $|{G_{w,x,y}}^{\Delta (x)}|$. However, Lemma 3.2 (iii) then implies t divides $| \textrm{PSL}_{n-2}(q)|$, contrary to t being a Zsigmondy prime. Therefore, Zsigmondy’s Theorem (see [2, 3.8]) implies that $q^{n-1} = 2^6$ or $q^{n-1}-1 = p^2-1$ and $p= 2^r-1$ is a Mersenne prime.

${\textbf{2}}^\circ$. We have $q^{n-1}\ne 2^6$.

Assume that $q^{n-1}= 2^6$. Then, as $n \ge 3$, we have one of the following cases $F^*(G_x^{\Delta (x)})\cong \textrm{PSL}_3(8)$ with $q_x= 8(8+1)$, $\textrm{PSL}_4(4)$ with $q_x= 4(4^2+4+1)$ or $\textrm{PSL}_7(2)$ with $q_x= 2(2^6-1)$. Since $s\ge 4$, $G_{x,y}$ has order divisible by $q_x^2$. The first case $|G_x^{[1]}|$ is coprime to 3 by Lemmas 2.7 and 3.3. Therefore, $|G_{x,y}|_3 =|G_{x,y}^{\Delta (x)}|_3=3^3< 3^4=(q_x^2)_3$, which is a contradiction. Similarly, the second case is impossible as $7^2$ does not divide $|G_{x,y}^{\Delta (x)}|$. Therefore, $F^*(G_x^{\Delta (x)})\cong \textrm{PSL}_7(2)$ and $G_x^{[2]}=Q_x$ is a 2-group by Lemma 3.3. Set $H=G_{w,x,y,z}$. Then, by Lemma 2.5 (v) $\pi (H)=\pi (G_{x,y})\supseteq \{2,3,5,7,31\}$ and by Lemma 2.4 (ii) $H\cap G_{x}^{[1]} \le G_z\cap G_{x}^{[1]}$ is a 2-group. Hence, $H^{\Delta (x)}\le G_{w,x,y}^{\Delta (x)} \cong 2^{10}: \textrm{SL}_5(2)$ and $\pi (H^{\Delta (x)}) \supseteq \{3,5,7,31\}$. By [8], the maximal over-groups of a Singer cycle in $\textrm{SL}_5(2)$ have order $5\cdot 31=155$ and so we conclude that $G_{w,x,y}^{\Delta (x)}= H^{\Delta (x)}O_2(G_{w,x,y}^{\Delta (x)})$. In particular, we see that H has a quotient isomorphic to $\textrm{SL}_5(2)$ and has order $2^{\ell }\cdot 3^2\cdot 5\cdot 7\cdot 31$ for some $\ell$. Since $s \ge 4$, H operates transitively on $\Delta (z) \setminus \{y\}$. Hence, for $a\in \Delta (z) {\setminus } \{y\}$, $|H:H_a|= q_x=126$ and so $|H_a|= 2^{\ell -1}\cdot 5\cdot 31$. Therefore,

$$\begin{aligned} 2^{10}.5.31\ge |H_a^{\Delta (x)}O_2(G_{w,x,y}^{\Delta (x)})/O_2(G_{w,x,y}^{\Delta (x)})|\ge 2^9\cdot 5\cdot 31>5 \cdot 31 \end{aligned}$$

and $H_a^{\Delta (x)}O_2(G_{w,x,y}^{\Delta (x)})/O_2(G_{w,x,y}^{\Delta (x)})$ contains a Singer cycle of $\textrm{SL}_5(2)$, which is a contradiction.

Because of $({\textbf{1}}^\circ )$ and $({\textbf{2}}^\circ )$, it remains to exclude the possibility that $q^{n-1}-1 = p^2-1$ with $p=2^r-1$ a Mersenne prime. Thus, $n=3$, $q_x = p(p+1) = 2^rp$ and $p-1=2(2^{r-1}-1)$ is not divisible by 4.

By Lemma 2.4 (ii),

$$\begin{aligned} G_{w,x,y,z} \cap G_x^{[1]}=G_{x}^{[1]} \cap G_z=G_x^{[2]} \end{aligned}$$

and by Lemma 3.3$|G_x^{[2]}|$ divides $(p-1)|Q_x|$. By Lemma 3.2$|G_{w,x,y,z}^{\Delta (x)}|$ divides $(p-1)^2p^2$. Hence $|G_{w,x,y,z}|$ divides $(p-1)^3p^2|Q_x|$. So, as 4 does not divide $p-1$, and $G_{w,x,y,z}$ acts transitively on $\Delta (z){\setminus }\{y\}$, $q_z=q_x=2^rp$ is not divisible by 16. Hence, $r \in \{2,3\}$. Furthermore, if $r=3$, then $|G_x^{[2]}|$ is even. We signal

${\textbf{3}}^\circ$. $q=p=2^{r}-1\in \{3,7\}$ and $|G_x^{[2]}|$ is even if $q=7$.

Suppose that $G_x^{[2]}$ has odd order. Then, $r=2$, $q=p=3$, $G_x=G_x^{[1]}L_x$, $L_x/Q_x \cong \textrm{PSL}_3(3)$ and $q_x=12$. By Lemma 2.7, $G_x^{[1]}$ also has odd order. Let $S \in {\textrm{Syl}}_2(G_{w,x,y,z})$. Then, $S \cap G_x^{[1]}=1$ and so $S\cong S^{\Delta (x)} \le G_{w,x,y}^{\Delta (x)}$. Hence, S is elementary abelian and $|S|\le 4$. Since $q_x=12$ and $q_x$ divides $|G_{w,x,y,z}|$ by Lemma 2.5 (v), we conclude that $|S|=4$. Since $s \ge 4$, and $q_z=q_x=12$, we now know that $|G_{w,x,y,z,a}|$ is odd for all $a \in \Delta (z)\setminus \{y\}$.

Let $T \in {\textrm{Syl}}_2(G_{y,z})$ with $S \le T$. Then, $T \cap G_z^{[1]}=1$ and $T \cong {\textrm{SDih}}(16)$ is semidihedral. Since S is elementary abelian of order 4, we have $Z(T) \le S$. In particular, $Z(T)G_{z}^{[1]}/G_{z}^{[1]}$ acts by conjugation inverting each element of $Q_yG_{z}^{[1]}/G_{z}^{[1]}$. Observe that $Q_y$ has 4 orbits of length 3 on $\Delta (z) {\setminus } \{y\}$ each of which is fixed by Z(T). Hence, Z(T) fixes a vertex $a \in \Delta (z) \backslash \{y\}$, contrary to $|G_{w,x,y,z,a}|$ being odd. This contradiction shows that $|G_x^{[2]}|$ is divisible by 2.

Assume that $|G_x^{[2]}|$ is even. Then, $q\in \{3,7\}$ by $({\textbf{3}}^\circ )$. Lemma 3.3 states that $G_x^{[2]}/Q_x$ is cyclic, and so the Sylow 2-subgroups of $G_x^{[2]}/Q_x$ have order 2. By Lemma 2.6, $C_y /Q_y$ is abelian and the Sylow 2-subgroups of $C_y$ are elementary abelian and have order at least 4. As $C_y ^{\Delta (x)}$ is normal in $G_{x,y}^{\Delta (x)}$ and $G_{x,y}^{\Delta (x)}/Q_y^{\Delta (x)}$ is isomorphic to a subgroup of $\textrm{GL}_2(q)$ containing $\textrm{SL}_2(q)$, Lemma 3.2(iv) yields that $C_y^{\Delta (x)}/Q_y^{\Delta (x)}$ is cyclic. Lemma 2.4 (ii) gives $G_x^{[1]} \cap C_y\le G_x^{[1]} \cap G_y^{[1]}=G_x^{[2]}$, and thus, $C_y^{\Delta (x)}/Q_y^{\Delta (x)}\cong C_y/G_x^{[2]}Q_y$. We deduce that $C_y/Q_y$ contains exactly three involutions. If $q_y >2$, then there exists $u, v \in \Delta (y)$ such that $G_u^{[2]}Q_y/Q_y \cap G_{v}^{[2]}Q_y/Q_y$ has an involution and this contradicts Lemma 2.5 (iv). Hence, $q_y=2$ and $G_y$ acts transitively on the three involutions in $C_y/Q_y$. Let $S_y \in {\textrm{Syl}}_2(C_y)$. Then, $S_yQ_y$ is normalized by $G_y$, and thus, $S_yQ_y$ has a unique $G_y$-conjugacy class of involutions.

Since $q_y=2$, we have $O^{p'}(G_{x,y}) \le G_y^{[1]}$ and so $O^{p'}(G_{x,y}) =O^{p'}(G_y^{[1]})$ is normal in $G_y$. Because $q=p\in \{3,7\}$, $O^{p'}(G_x)/Q_x \cong \textrm{PSL}_3(p)$ or $\textrm{SL}_3(p)$ and, as the Schur multiplier of $\textrm{PSL}_2(p)$ has order 2, we get $O^{p'}(G_y^{[1]})/Q_x \cong \textrm{ASL}_2(p)$. In particular, $G_x^{[2]} \cap O^{p'}(G_y^{[1]})=Q_x$.

Let $a \in O^{p'}(G_y^{[1]})$ be an involution and set $T=C_y\langle a\rangle$. Then, Lemma 3.2 (iv) implies

$$\begin{aligned}{}[T^{\Delta (x)}, O^{p'}(G_y^{[1]})^{\Delta (x)}]= [a^{\Delta (x)},O^{p'}(G_y^{[1]})^{\Delta (x)}] [C_y^{\Delta (x)}, O^{p'}(G_y^{[1]})^{\Delta (x)}]\le Q_y^{\Delta (x)} \end{aligned}$$

and so $T\le C_yG_x^{[1]}$, as $|C_y^{\Delta (x)}|$ is even. Hence, using Lemma 2.4 (i)

$$\begin{aligned} T = C_yG_x^{[1]}\cap T= C_y( G_x^{[1]}\cap T)\le C_y (G_x^{[1]}\cap G_y^{[1]})=C_yG_x^{[2]}=C_y. \end{aligned}$$

Thus, $a\in C_y\cap O^{p'}(G_y^{[1]})$. Since $O^{p'}(G_y^{[1]})$ is normal in $G_y$, we deduce that $S_y \le O^{p'}(G_y^{[1]})$ which is impossible as $O^{p'}(G_y^{[1]})\cap G_x^{[2]}= Q_x$. This completes the proof. $\square$

Lemma 3.5

Suppose that $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$. Pick $U \in {\textrm{Syl}}_3(G_y)$ and set $D= U\cap C_y$ and $F =D \cap L_{x,y}$. Then

• (i) $q_x=6$ and $q_y=2$;

• (ii) $|G_x^{[2]}/Q_x|=3$;

• (iii) D is elementary abelian of order 9, $C_y= DQ_y$ and $D \in {\textrm{Syl}}_3(G_y^{[1]})$;

• (iv) F has order 3, $Q_yF \trianglelefteq G_y$, $F=Z(U)$ and U is extraspecial of order 27.

Proof

As $G_x^{\Delta (x)} \cong \textrm{PSL}_3(2)$, we have $p=2$, $q_x=6$ and Lemma 3.4 implies $|G_x^{[2]}/Q_x|=3$.

By Lemma 2.6, $C_y/Q_y$ is an elementary abelian 3-group of rank at least 2. Since $G_{x}^{[1]} \cap G_{y}^{[1]} \le G_{x}^{[2]}$ and $G_{x,y}^{\Delta (x)} \cong \textrm{Sym}(4)$, we deduce $C_y/Q_y$ has order 9. Hence, D is elementary abelian of order 9, $C_y= DQ_y$ and $D \in {\textrm{Syl}}_3(G_y^{[1]})$. This proves (iii).

We know $FQ_y$ has index 3 in $C_y$ and $FQ_y$ is normalized by $G_{x,y}$. If $FQ_y= G_{u}^{[2]}Q_y$ for some $u \in \Delta (y){\setminus }\{x\}$, then as $G_{x,y}$ is transitive on $\Delta (y)\setminus \{x\}$, we have $G_{u}^{[2]}Q_y=G_{z}^{[2]}Q_y$ for all $u \in \Delta (y)\setminus \{x\}$ contrary to Lemma 2.5 (iv). So of the four subgroups of index 3 in $C_y$, there are only three candidates for $G_{u}^{[2]}Q_y$ and so we conclude that $q_y=2$ and this proves (i).

Because U acts transitively on $\Delta (y)$, U permutes the three subgroups of $\{G_u^{[2]} Q_y \mid u \in \Delta (y)\}$. In particular, as $D \in {\textrm{Syl}}_3( G_y^{[1]})$ and $G_y/G_y^{[1]}\cong \textrm{Sym}(3)$, we have U is non-abelian of order 27. As $FQ_y$ is normalized by U, $F = Z(U)$. This concludes the proof. $\square$

Lemma 3.6

Assume that $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$. Then, $L_x$ has either 1, 2, 3 or 6 non-central $L_x$ chief factors, each of which is 3-dimensional.

Proof

We establish [10, Hypothesis] with $p=2$ using boldface letters for the groups used in [10, Hypothesis]. So set ${\textbf{M}}= L_x$, ${\textbf{E}}= Q_x$, ${{\textbf {B}}}\in {\textrm{Syl}}_2({\textbf {M}})$ and $\textbf{P}_1, \textbf{P}_2\le \textbf{M}$ such that $\textbf{P}_1 \cap \textbf{P}_2=\textbf{B}$. To reassure ourselves, this means that ${\textbf{M}}/{\textbf{E}}=L_x/Q_x \cong \textrm{PSL}_3(2)$, ${{\textbf {B}}}/{{\textbf {E}}} \cong {\textrm{Dih}}(8)$ and ${\textbf{P}_1}/{\textbf{E}}\cong {\textbf{P}_2}/{\textbf{E}}\cong \textrm{Sym}(4)$. We choose notation so that ${\textbf{P}_1}=L_{x,y}$. We have $\textbf{P}_1=\textbf{P}_1^*$ and $\textbf{P}_2=\textbf{P}_2^*$. This means that [10, Hypothesis (WBN)] is satisfied. So we take $\textbf{S}=\textbf{B}$ and ${\textbf{T}}= Q_y$. By Lemma 2.8, [10, Hypothesis (P)] holds. Since $\textbf{P}_1=\textbf{P}_1^*$ and $\textbf{P}_2=\textbf{P}_2^*$, $O^{2'}(\textbf{P}_1)= \textbf{P}_1$ and $O^{2'}(\textbf{P}_2)= \textbf{P}_2$. Thus, setting

$$\begin{aligned} \textbf{L}=\langle O^2(O^{2'}(\textbf{P}_1^*)),O^2(O^{2'}(\textbf{P}_2^*))\rangle = \langle O^2(\textbf{P}_1^*),O^2(\textbf{P}_2^*) \rangle \end{aligned}$$

and remembering $\textbf{E}\le \textbf{T}$, we have

$$\begin{aligned} \textbf{L}\textbf{E}= \textbf{L}\textbf{T}. \end{aligned}$$

So, as $\textbf{E}$ is a 2-group, $\textbf{E}/ O_2(\textbf{E})$ is a $2'$-group, and $\textbf{E}\le \textbf{FO}_2(\textbf{E})$ for every subgroup $\textbf{F}\le \textbf{M}$ with

$$\begin{aligned} \textbf{L}\le \textbf{E}\textbf{F}. \end{aligned}$$

Hence, [10, Hypothesis (A)] and [10, Hypothesis (B)] both hold.

Since [10, Hypothesis] holds, and since ${\textbf{M}}= L_x$, we can conclude from [10, Theorem] that one of the cases (4), (5), (8), (11) or (13) of that theorem holds. In particular, we see that ${\textbf{E}}= Q_x$ has either 1, 2, 3 or 6 non-central $L_x$ chief factors, each of which is 3-dimensional. This proves the claim. $\square$

Lemma 3.7

• (i) $L_x$ has two non-central chief factors in $Q_x$, each of which is 3-dimensional.

• (ii) F has three non-central chief factors in $Q_y$ (where F is as in Lemma 3.5).

Proof

Consider the non-central chief factors of $G_y$ in $Q_y$. If the chief factor is not centralized by $F= Z(U)$, then, as U is extraspecial of order 27, the chief factor has order a multiple of $2^6$ and F acts fixed point freely. Thus, the number of F-chief factors is a multiple of 3 which we denote by 3f. As $G_y$ has characteristic p, $f \ge 1$.

From the perspective of $L_x$, we have $Q_y/Q_x$ is a non-central F-chief factor, and, as each $L_x$ non-central chief factor in $Q_x$ is 3-dimensional by Lemma 3.6, $FQ_x$ has one non-central chief factor for each $L_x$ non-central chief factor. Thus, $L_x$ has $3f-1$ non-central chief factors in $Q_x$. Using Lemma 3.6, we deduce that $L_x$ has 2 non-central chief factors in $Q_x$. Thus, (i) holds and (ii) follows from this. $\square$

Recall the definition of $Z_x$ and $Z_y$ from Notation 2.1 and observe $Z_y \le Z_x$, since $\Delta$ is of local characteristic 2 and $Q_x \le Q_y$.

Lemma 3.8

Suppose $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$. Then

• (i) $Q_y=Q_xQ_z$ and $Z_x \cap Z_z=Z_y$;

• (ii) $[Z_x, F] \ne 1$.

Proof

We continue the notation from Lemma 3.5.

(i). We know $FQ_y \le G_y^{[1]}$. Since F normalizes $Q_z$ and $Q_z \ne Q_x$, we have $Q_xQ_z= Q_y$. Since $\Delta$ is of local characteristic p and $Q_y=Q_xQ_z$, we glean $Z(Q_y) \le Z(Q_x) \cap Z(Q_z) \le Z(Q_y)$.

(ii). Suppose $[Z_x, F]=1$. Then $[Z_x, O^2(L_x)] = 1$ and $L_x=Q_xO^2(L_x)$. Since $O^2(L_x)$ acts transitively on $\Delta (x)$ and $Z_y\le Z_x$, we obtain $Z_y=1$ from Lemma 2.2 and this is a contradiction.$\square$

Lemma 3.9

Suppose $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$. Then, $Z_z\not \le G_x^{[1]}$.

Proof

Again we use the notation started in Lemma 3.5. By Lemma 3.4, we know $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$. Set $V_y = \langle Z_u\mid u \in \Delta (y)\rangle$. Then, $V_y\le Q_y$ is normalized by $G_y$ and F does not centralize $Z_x\le V_y$ by Lemma 3.8 (ii). Hence, U acts faithfully on $V_y$ and so we conclude that F has at least three non-central chief factors in $V_y$. Since F has exactly three non-central chief factors in $Q_y$ and $Q_y/Q_x$ is such a factor, we conclude that $V_y \not \le G_x^{[1]}$ and this delivers the claim. $\square$

Proof of Proposition 3.1

Lemma 3.4 gives $G_x^{\Delta (x)}\cong \textrm{PSL}_3(2)$ and, by Lemma 3.9, $Z_z \not \le Q_x$. As $F\le G_y^{[1]}$ normalizes $Z_z$, $Q_y=Q_xZ_z=Z_xQ_z$. Since $Q_x\cap Z_z$ is centralized by $Z_xQ_z$, we deduce $Q_x \cap Z_z=Z_y$.

Let $u \in \Delta (y) \setminus \{x,z\}$. Since $s \ge 4$, we have $Z_z \not \le Q_{u}$. We calculate using the fact that $Q_x$ is the unique Sylow 2-subgroup of $G_{x}^{[2]}$ by Lemma 3.5 (ii)

$$\begin{aligned}{}[G_{x}^{[2]},Z_z]\le G_x^{[2]} \cap Z_z \le Q_x \cap Z_z=Z_y \end{aligned}$$

and

$$\begin{aligned}{}[G_u^{[2]},Z_z]\le Q_{u} \cap Z_z=Z_y. \end{aligned}$$

It follows that

$$\begin{aligned}{}[F,Z_z] \le [G_u^{[2]}G_x^{[2]},Z_z] \le Z_y\le Q_x. \end{aligned}$$

However, $Z_zQ_x/Q_x$ is not centralized by F, and so this is impossible. This contradiction completes the proof. $\square$

4 The main theorem

Suppose that X is a 2-transitive group in its action on $\Omega$. Then, [1, Lemma 2.2] (for example) yields that either there is a prime r such that $F^*(X)$ is a regular elementary abelian r-group, or $F^*(X)$ is a non-abelian simple group. In the first case, we say that X is of regular type, and in the second that X is of simple type. When X is of simple type, the description of $F^*(X)$ and $\Omega$ is conveniently presented in [1, Lemma 2.5] (this result requires the classification of the finite simple groups). Since $G_x^{\Delta (x)}$ acts 2-transitively on $\Delta (x)$ and as we also know that $1\ne Q_y^\Delta (x) \unlhd G_{x,y}^{[1]}$, Proposition 3.1 combined with [1, Lemma 2.5] yields

Lemma 4.1

The group $F^*(G_x^{\Delta (x)})$ is either of regular type or is of simple type and is isomorphic to a rank 1 group of Lie type in characteristic p in its natural permutation representation (including $\textrm{Ree}(3)^\prime$).

$\square$

We can now move directly to the proof Theorem 1.1.

Proof of Theorem 1.1

Set $N= O^p(L_x)Q_x$. Then $O_p(N)= Q_x$ and $N= O^p(N)Q_x$. Define $S= Q_y$ and $\widetilde{G}= G_x/Q_x$. Then, with $L= L_x$, we have $L= NS$, $O_p(N)= O_p(L)=Q_x < Q_y=S$ and $C_S(Q_x) \le C_G(Q_x)\le Q_x$. Furthermore, [1, Hypothesis 3.3(b) and (c)] follows from Lemma 2.3(v) and Lemma 4.1, respectively. Because of Lemma 2.8 and [1, Lemma 3.8], we are in the same conclusions as [1, 7.7 steps $\textbf{1}^\circ$ and $\textbf{2}^\circ$]. Following [1, 7.7 steps $\textbf{3}^\circ$ through $\textbf{9 }^\circ$] verbatim (being careful to note the role of x and y are reversed) yields $O^p(L_x) \cong \textrm{AGL}_2(q)'$, $q=p^r$ with p odd, and $Q_x$ elementary abelian. This completes the proof. $\square$